Answer: MN = 15
\[9 \times A(\triangle PQR) = 16 \times A(\triangle LMN)\] \[\therefore \frac{A(\triangle LMN)}{A(\triangle PQR)} = \frac{9}{16}\]Since \(\triangle LMN \sim \triangle PQR\), MN and QR are corresponding sides, so by the theorem of areas of similar triangles,
\[\frac{MN^2}{QR^2} = \frac{9}{16}\] \[\frac{MN}{QR} = \frac{3}{4}\] \[MN = \frac{3}{4} \times 20 = 15\]